Find a vector equation and parametric equations for the line. The line through the point (0, 14, -10) and parallel to the line x=−1+2t, y=6−3t, z=3+9t

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Find a vector equation and parametric equations for the line. The line through the point (0, 14, -10) and parallel to the line x=−1+2t, y=6−3t, z=3+9t

Juan Spiller · Accepted Answer

x=−1+2t, y=6−3t, z=3+9t The direction vector is obtained from the coefficient of t. v=&amp;lt;2,−3,9&amp;gt; If the position vector pointing at the point (0,14,−10) is r0=0,14,−10&amp;gt;, then the vector equation of the line is r=r0+tv =&amp;lt;0,14,−10&amp;gt;+t&amp;lt;2,−3,9&amp;gt; We can also write it in more tedious &quot;ijk&quot; form: r=(0i+14j−10k)+t(2i−3j+9k) r=(14j−10k)+t(2i−3j+9k) The parametric equations form of a line use x=x0+at y=y0+bt z=z0+ct where &amp;lt;a,b,c≥v=&amp;lt;2,−3,9&amp;gt; from before, and (x0,y0,z0)=(0,14,−10), the point on the line x=0+2t y=14−3t z=−10+9t

stomachdm · Accepted Answer

The answer is the same as in the book. Thanks!

Find a vector equation and parametric equations for the line.

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